Summary Of Ordinary Differential Equations 4u205h

Up to Midterm 1 First order linear ODE y   p(t ) y  g (t ) Exact Differential M (t , y )  N (t , y ) y   0 M N  Exact if t y Implicit solution    M (t )dt  h( y ) such that

   M, N t y

Integrating factors Make it so M and N become related by the product rule  M   Ny '  ( yO) If p is integrable then,   e  p (t ) dt  N M   N     y t y   t If it is known that μ is a function of only y or t, then simplify First order linear ODE solution ( y (t )  (t ))  g (t )  (t )  F  In general, M

y (t )  (t )  F t

F (t )  F (t0 )    (t ) g (t )dt  t0

Second order homogenous ay  by  cy  0 Characteristic equation (assume solution is in the form of y  Ce rt ar 2  br  c  0 with roots r1 , r2 rt rt Solution y  C1e 1  C2e 2 rt rt Double root: y  C1e  tC2 e

t Complex conjugate roots: r     i, y  e (C1 cos(  t )  C2 sin(  t )) Wronskian matrix (Second order) f g W ( f , g )(t )  f  g If f and g are linearly independent, W  0 for all t For two solutions y1 , y2 to y   py  gy  0  pdt W  Ce  Higher order: Nth order: rows are f, g, h, etc, and columns are D0, D1, D2, etc. Up to Midterm 2 Higher-order homogenous DE with constant coefficients (very similar to 2nd-order) Find roots to characteristic equation Complex will come in pairs, do cos/sin thing Multiple roots, multiply by increasing powers of t Non-homogeneous second order DE Solution is sum of steady-state (particular) and transient (complementary) solution Transient is solution of homogenous equation Steady-state is the solution to the non-homogenous equation, unique to IVCs Solving for steady-state solutions Reduction of order Have one solution Guess y2 (t )  u (t ) y1 (t ) , differentiate, plug in, find u

Variation of Parameters (Second-order) Homogenous solutions y1, y2 to y   py  qy  g Guess solution Y  u1 y1  u2 y2 and let u1 y1  u2 y2  0 , differentiate Y   u1 y1  u2 y2 and Y   u1 y1  u2 y2  u1 y1  u2 y2 Substitute and get u1 y1  u2 y2  g t t y2 y1 gdt   y2  gdt  Solution Y  Y0   y1 t0 t 0 W(y , y ) W ( y1 , y2 ) 1 2 Larger order (second order, too) Nth order has N parameters (u1,2…N) N

Y   yiui i 1

Wi gdt  W Wi is the Wronskian with the ith column replaced by [0,0,…1] Undetermined coefficients Guess solution’s form, find arbitrary coefficients t g  e t , guess y p  Ae g  sin( t ) , guess y p  A sin( t )  B sin( t ) ui  

Non-constant coefficients Series Solution (for a neighborhood near an ordinary point a [ p (a ) y, p( a)  0 ]) 

i Solution is of the form y   ai t n 0





n 1

n2

i 1 i 2 Derive to get y    iait , y   i (i  1)ai t

Substitute into the equation, try to combine Create similar exponents for t (usually i) Shift summations: n’=n+2 for instance Ex: constant coefficients 



n 0

n 0

y    (i  1)ai 1t i , y    (i  2)(i  1)ai  2t i Then, sum of power series is zero, Theorem: all coefficients zero Solve for coefficients (get a recursive formula) Unknown starting coefficients(a0, a1, etc) IVC coefficient 

i Find solution by plugging coefficients into y   ai t n0

Up to Final Cauchy-Euler Equation n

a y i 0

i

(i )

x i  0, ax 2 y   bxy   cy  0, an  1

Assume solution is in the form y  x r Get a polynomial find roots r ri Real r: y   ci x , multiply by ln(x) for multiple roots i

Complex r: given x r  er ln x  x  i   x  ei  ln x y | x | (c1 cos(  ln | x |)  c2 sin(  ln | x |)) Can shift x by x0 (singular point) by x-x0 into x’ Singular Points

py  qy   ry  0 , x0 singular if p(x0)=0 q r x  x0 ) , lim( x  x0 ) 2 finite (they are “analytic”), else irregular Regular if xlim(  x0 x  x 0 p p Solutions near a Regular Singular Point 



n 0

n 0

n 2 n xq/p and x2r/p analytic at x=x0, so ( x  x0 ) p   pn ( x  x0 ) , ( x  x0 ) q   qn ( x  x0 )

Take L[y], multiply by x and divide by p to get x 2 y  x( xp ) y  ( x 2 q ) y  0 2 If you take p0 and q0 only: x y  xp0 y  q0 y  0 , an Euler equation (indicial equation) 2







r n r  n 1 , y    an (r  n)(r  n  1) x r  n  2 Assume solution y   an x , y   an (r  n) x n0

n 0 r+n

n 0

r

Get some power series in x as well as an x term, the same as the n=0 equation case xr coefficient=0 is the indicial equation, roots are exponents at the singularity Solve in order highest to lowest since usually lower root is more complicated Solve lower roots only if they do not differ by an integer (else complicated) Combine sums, coefficient=0 yields a series equation, solve for each root from the indicial equation Laplace Transformation 

L{ f (t )}  F ( s )   e  st f (t )dt 0

at Exists if f (t )  Ke for t>M; a, K, M, constants, K>0, M>0 on s>a Laplace transformation is linear L{ f (t )}  sL{ f (t )}  f (0) n

L{ f ( n ) (t )}  s n L{ f (t )}   s n i f ( i 1) (0) i 1

Uses of the Laplace Transformation L{L[y]} gives a function Y(s)=L{y} which is the Laplace transformation of the solution Y(s) can be analyzed for solutions by doing a Laplace transformation backwards Protips: Partial fractions for quotient polynomial Y(s) 1 a s L{e at }  , L{sin at}  2 , L{cos at}  2 2 sa s a s  a2 Textbook p317 for Laplace transformations of elementary functions equation